Math For Accounting S5 SB
Math For Accounting S5 SB
Senior
                5
         for Rwandan SchoolsStudent   Book
     EXPERIMENTAL VERSION
         © 2023 Rwanda Basic Education Board
                   All rights reserved
 This book is the property of the Government of Rwanda.
Credit must be provided to REB when the content is quoted
FOREWORD
Dear Student,
Rwanda Basic Education Board (REB) is honoured to present Senior 5
Mathematics book for the students of Accounting Profession Option which
serves as a guide to competence-based teaching and learning to ensure
consistency and coherence in the learning of the Mathematics. The Rwandan
educational philosophy is to ensure that you achieve full potential at every level
of education which will prepare you to be well integrated in society and exploit
employment opportunities.
The government of Rwanda emphasizes the importance of aligning teaching
and learning materials with the syllabus to facilitate your learning process.
Many factors influence what you learn, how well you learn and the competences
you acquire. Those factors include the relevance of the specific content, the
quality of teachers’ pedagogical approaches, the assessment strategies and
the instructional materials available. In this book, we paid special attention
to the activities that facilitate the learning process in which you can develop
your ideas and make new discoveries during concrete activities carried out
individually or in groups.
In competence-based curriculum, learning is considered as a process of active
building and developing knowledge and meanings by the learner where
concepts are mainly introduced by an activity, situation or scenario that helps
the learner to construct knowledge, develop skills and acquire positive attitudes
and values.
For efficiency use of this textbook, your role is to:
      – Work on given activities which lead to the development of skills;
      – Share relevant information with other learners through presentations,
         discussions, group work and other active learning techniques such
         as role play, case studies, investigation and research in the library, on
         internet or outside;
      – Participate and take responsibility for your own learning;
      – Draw conclusions based on the findings from the learning activities.
To facilitate you in doing activities, the content of this book is self-explanatory
so that you can easily use it yourself, acquire and assess your competences. The
book is made of units as presented in the syllabus. Each unit has the following
structure: the unit title and key unit competence are given and they are followed
by the introductory activity before the development of mathematical concepts
that are connected to real world problems more especially to production,
finance and economics.
MURUNGI Joan
 Unit 2: Differentiation/Derivatives...............................................................................40
       2.1 Differentiation from first principles.................................................................41
Key Unit competence: Use matrices and determinants notations and properties
                       to solve simple production, financial, economical,
                       and mathematical related problems.
Introductory activity
  The table below shows the revenue and expenses (in Rwandan francs) of
  a family over three consecutive months:
      1 
       
Thus,  3  is a column matrix.
      6
       
– A square matrix is a matrix in which the number of rows is equal to the
  number of columns; that is, matrix A of order n × p is a square matrix if and
  only if n = p ;
In this case, instead of saying a matrix of order n × n , we, sometimes, simply say
a matrix of order n .
respectively.
If A = ( aij ) is a square matrix of order 2 , then i and j assume values in the set
                              a      a12 
  {1, 2}.Therefore, ( aij ) =  11        .
                               a21   a22 
  In the same way, if A = ( aij ) is a square matrix of order 3 , then i and j assume
                            1 3 4 
                                      
  In the matrix =M (= aij )  5 12 13  ,
                            7 6 0 
                                      
         a) Write down the value of a23 and the value of a31
         b) Explain why the matrix is a square matrix
  Solution:
        a) The entry on the second row and third column is a23 = 13 ;
            The entry on the third row and first column is a31 = 7 .
         b) M is a square matrix because it has the same number of rows and
            columns
                 8 6 2                                      7
         a) A =               b) B =   ( −2   1 3)   c) C =  
                 −1 1 0                                      −5 
    2.   A shoe shop sells shoes for men and ladies. The first week, it sold 7
         pairs of men’s shoes and 15 pairs of ladies’ shoes. The second week,
         it sold 9 pairs of ladies ‘shoes and 4 pairs of men’s shoes. Record
         this information as a 2 × 2 matrix, stating what the rows stand for,
         and what the columns stand for.
Situation1:
A class consists of boys and girls who are boarders or day scholars. The
class teacher records the data by the matrix
      8 9
 A=        , where the columns represent the numbers of boys and girls,
     17 6 
and the rows represent the numbers of boarders and day scholars.
Situation2:
Two brothers sell shirts and shoes, in two different shop I and II, for two
consecutive weeks. The Elder brother records his data by the matrix
      8 9
 B=         , where the columns represent the numbers of shirts and
     17 6 
shoes, and the rows represent the numbers of items sold in week1, and in
week2.
                                                               8 9
The younger brother, also, records his data by the matrix C =        ,
                                                              17 6 
where the columns represent the numbers of shirts and shoes, and the rows
represent the numbers of items sold in week1, and in week2. Comment on
the following, for matrices A, B and C:
CONTENT SUMMARY
             A = ( aij )
Two matrices             and B = ( bij ) are equal if and only if:
   i) they have the same order;
   ii) the corresponding entries (that is the entries occupying the same position,
       in terms of rows and columns) are equal.
   iii) The nature of the entries in the two matrices is the same.
Note: When discussing the equality of matrices, we assume the nature of the
entries in the two matrices to be the same.
                   3          0            2 +1 7 − 7 
                                                      
                              −18             13
Thus, matrices A=  −1              and B =         2  are equal.
                             −9            −13        
                                                  8 
                   16         23           4
Example1.1.2.
                         2x −1  −3              9 −3 
Given that matrices A =                and B =        are equal, find the
values of x and y .      2     2x + y           2 −11
Solution:
Matrices A and B are both of order 2 × 2 and for the corresponding entries to
be equal, we have:
 2 x − 1 =9                                        x=5
              . Solving simultaneously, we get    
2 x + y = −11                                      y = −21
                            1 1            5 − 4 3 
             1 1                                             1    23 − 5 
     a) A =      and B = 1 1      b) A =        12  and B =              
             1 1          1 1
                                              2                 9 −1   4 
                                                  3
                                 2       1
                                  2x − x              1            y
     2. Given that matrices X =           3  and Y =  2               are equal,
                                                   t + t        −5 
                                  0      z 
     6 5 13           7 8 2 
 S =          and T =          .
    11 13 10           7 17 20 
            a) Write down the order of each of the two matrices S and T. How are
               these two orders?
            b) Determine a single matrix for the total sales for this retailer for
               the last three weeks in the two shops.
            c) Predict the conditions for two matrices to be added and how to
               obtain the sum of two matrices.
 a11   a12   a13   b11 b12     b13   a11 + b11    a12 + b12   a13 + b13 
                                                                         
 a21   a22   a23  +  b21 b22   b23  =
                                         a21 + b21    a22 + b22   a23 + b23 
a      a32   a33   b31 b32   b33   a31 + b31
                                                      a32 + b32   a33 + b33 
 31
Example 1.2.1.
                   11 29           6 34 
Given matrices P =        and Q =        , find the matrices:
                    7 14          3 7 
     a) Q − P
b) P + Q
Solution:
                6 34  11 29   6 − 11 34 − 29   −5 5 
        a) Q=
            −P       − =                 =          
                3 7   7 14   3 − 7 7 − 14   −4 −7 
          3 5    10 3 5 
     a) A =
     =        ; B          
          8 1     11 11 4 
           1 17         11 29 
=  b) A =         ; B          
          14 3          8 6
2. In a sector of a district, there are three secondary schools, A, B and C
    having both boarding and day sections for both boys and girls. The
    distribution of the students in the three schools are given, respectively
CONTENT SUMMARY
A matrix can be multiplied by a specific number; in this case, each entry of the
matrix is multiplied by the given number. This type of multiplication
                                                                       is called
scalar multiplication, since the matrix is multiplied by a single real number,
and real numbers are also called scalars.
           A = ( aij )                                         =      ( aij )
                                                                 α A α=            (α a ) ,
Thus, if                 of any order n × p and α a scalar, then                       ij
Example 1.2.2.
                    1 −2          6 3
Given matrices A =        and B =     , find the matrix 2 A + 3B
                   7 4            2 7
           1 −2   6 3   2 −4  18 9   20 5 
2A =
   + 3B 2        + 3 =       + =          
           7 4   2 7  14 8   6 21  20 29 
                              3 4          −2 1 
      1.   Given matrices A =     and B =        ,find:
                              5 0          5 3
           a) 4 A − 5 B
         b) 2( A + B)
      3) 2. In November a shop sells clothes and shoes for both boys and
         girls.
            The number of clothes and shoes sold for boys and girls are
                                        15 21 
            recorded in the matrix N =         , where the number of boys
            and girls are in            19 28 
                                            1 3 2
 or two supermarkets S1 and S2. Matrix M =             shows the number
                                             2 4 3
 of kilograms of each of the three items(sugar, rice and beans)bought by
                                      1800 1600 
                                                 
 each of the two friends .Matrix P =  2500 2000  shows the price per
                                      1500 1200 
                                                 
 kilogram, in Rwandan francs, of sugar, rice and beans in supermarkets S1
 and S 2 , respectively.
            CONTENT SUMMARY
            Let A and B be matrices of order n × p , and m × r , respectively. Matrices A and
            B can be multiplied, in this order, if and only if p = m , that is, the number of
            columns of the first matrix is equal to the number of rows of the second matrix.
            In this case, we say that matrices A and B, in this order, are conformable for
            multiplication. The product A × B is of order n × r , that is the product A × B
            has the same number of rows as matrix A , and the same number of columns
            as matrix B.
A: n× p
B: p×r
A.B : n × r
              2.   Calculate the entries of the product as follow: Let A.B = ( cij ) . To determine
                   the entry cij , multiply the entries along the ith row of the first matrix by
                   the corresponding entries down the jth column of the second matrix and
                   add the products.
                                  a    a12           b11 b12 
            In particular, if A =  11       and B =           , then
                                   a21 a22           b21 b22 
                                                 3 1
                              2 3 0                 
  1.   Consider matrices A =          and B =  −2 4 
                              −1 4 1           2 −1
                                                      
Solution:
      a)
A: 2×3
B: 3× 2
       A.B : 2 × 2
 A and B are conformable for multiplication, since the number of columns of A
is equal to the number of rows of B
In the same way,
B : 3× 2
A: 2×3
       B.A : 3 × 3
 B and A are conformable for multiplication, since the number of columns of B
is equal to the number of rows of A.
       b) The order of the product A.B is 2 × 2 , and the order of the product
          B. A is 3 × 3 .
Multiplication of matrices is not commutative. In general, for matrices A and
B, A.B ≠ B. A
                                               5
                               1 3 −1         
   2.   Consider matrices A =          , B =  2  and=
                                                        C           (2   −3) .
                               −2 5 3         4
                                                
Solution:
                    5
         1 3 −1    7 
=a) A.B =        . 2   
         −2 5 3   4  12 
                     
                      7           14 −21 
                .B).C   . ( 2=
             ( A=              −3)         
                      12          24 −36 
                          10 −15 
              1 3 −1             14 −21 
    A.(B.C ) 
    =               =  .  4 −6           
              −2 5 3   8 −12   24 −36 
                                 
 We can predict that multiplication of matrices is associative, that is, for all
 matrices A, B and C , conformable for multiplication, ( A.B).C = A.( B.C )
Solution:
                1 1  −1 1   0 0 
=     a) A.B =       .               
                1 1  1 −1  0 0 
A matrix, in which all the entries are zeros is said to be the null matrix or the
zero matrix.
 For matrices, the equality A.B = 0 does not imply A = 0 or B = 0 , that is , the
 product of two matrices can be the null matrix, yet none of the factors is a null
 matrix.
a) A = ( 3 1 12 ) and B = ( 4 2 1)
                                   3 2
           b) A = ( 5 2 ) and B =       
                                  7 6
        2. A company’s input requirement over the next two months for two
           inputs X and Y are given (in numbers of units of each input) by the
                        2 1
          matrix M =        .
                        6 8
    The company can buy these inputs from two suppliers, whose prices for
                                                4 2
    the two inputs are given by the matrix N =      , where the two rows
                                               5 1
    represent the suppliers and the columns represent the prices.
    Obtain the matrix for the total input bill for the next two months for both
    suppliers.
                      1 0 2            3 2 2 
                                     1          
              =
  Consider matrices A  0 1 −3= and B 5  3 2 −3 
                       1 −1 0           1 −1 −1 
                                                
        a) Find the product A.B , and write down its order.
        b) Describe the entries of the product.
        c) If such a matrix is called identity matrix, which of the following
                                      1 1                    1 0        0 1
           are identity matrices: i)          ii) (1)   iii)        iv)    
                                      1 1                    0 1        1 0
              1 0
                   
           v)  0 1 
              1 0
                   
        d) Matrix B is said to be the inverse of matrix A. When do we say that
           matrix X is inverse of matrix M?
CONTENT SUMMARY
    – A square matrix with each element along the main diagonal (from
      the top left to the bottom right) being equal to 1 and with all other
      elements being 0 is said to be the identity matrix, it is denoted by I;
For any square matrix A of order n × n , and the identity matrix I of order n × n
, we have:
A.I = A and I .A = A , that is, I is the identity element for multiplication of
matrices.
Example 1.2.4.
                                                                   1 2 
Use elementary row operations to find the inverse of matrix A =          , and
                                                                   1 −3 
verify that the product of the matrix A and its inverse is the identity matrix.
Solution:
                              1 2 1 0 
Consider the augmented matrix          
                              1 −3 0 1 
1 2 1 0                        1 2      1 0
               R2 → R2 − R1                ;
1 −3 0 1                          0 −5 −1 1 
                                 3 2
1 2 1 0              2    1 0
            R1 → R1 + R2       5 5;
 0 −5 −1 1           3              
                            0 −5 −1 1 
                                    3 2 
       3 2                    1 0 5 5 
1 0 5 5  R → − 1 R                      ;
            2 5
                    2
                                0 1 1 − 1 
  0 −5 −1 1                             
                                    5   5
                                     3 2 
                                      5 5  1 3 2 
                           =
Therefore, the inverse matrix is A−1 =               
                                      1 − 1  5  1 −1
                                             
                                     5    5
                       1 2  1  3 2  1 1 2   3 2 
The=product is A. A−1 =      .                 .   
                       1 −3  5  1 −1 5 1 −3   1 −3  				
                  1 5 0 1 0
      = =                       
		 5  0 5   0 1 
                       1 1 1 
                             
  Given the matrix M = 1 2 3  ,
                       1 3 2 
                             
          a) Write down the identity matrix I of order 3 × 3 and determine
             the augmented matrix ( M / I )
                                          4 1          3 2
          following square matrices A =       and B =      :
                                          5 3         12 8 
          a) In matrix B, how are the entries in row 2 obtained from the
             corresponding entries in row 1?
          b) Can you suggest a means for characterizing a square matrix as
             singular or not?
                                       4 3 7
                                              
     2.   Determine whether matrix A =  1 6 7  is singular or not.
                                        3 1 4
                                              
                                                                               a c
is the unique real number denoted and defined by det =
                                                     A                           = ad − bc .
                                                                               b d
        a a ' a '' 
                   
If A =  b b ' b ''  is a square matrix of order 3 × 3 , then the determinant of A
        c c ' c '' 
                   
                                                         a a ' a ''
is the unique real number denoted and defined by det A = b b ' b ''
                                                         c c ' c ''
= (ab ' c ''+ a ' b '' c + a '' bc ') − (cb ' a ''+ c ' b '' a + c '' ba ') : a sum of six terms, each term
having three factors
The following techniques can be used in evaluating the determinant of a
                   a a ' a '' 
                              
square matrix A =  b b ' b ''  of order 3 × 3 :
                   c c ' c '' 
                              
Sarrus’ method
  1.    Copy the first two columns as the fourth and fifth columns:
  2.    Multiply the entries along the descending arrows and add the products
        to obtain P1
      a a ' a ''
det A =
      b b ' b '' =       (ab ' c ''+ a ' b '' c + a '' bc ') − (cb ' a ''+ c ' b '' a + c '' ba ')
                 P1 − P2 =
      c c ' c ''
Expansion by cofactors
We have:
= a11 (a22 a33 − a32 a23 ) − a12 (a21a33 − a31a23 ) + a13 (a21a32 − a31a22 )
The determinant of the matrix remaining after deleting the row and the column
of an entry is called the minor of that element. Thus, if M ij is the matrix
remaining after deleting the i th row and the j th column of a square matrix
Example 1.3.1.
                                                 3 5 1
                                                        
1.   Calculate the determinant of the matrix A =  1 2 4  by expanding along:
                                                  4 3 2
       a) The first row                                 
       b) The second column
Solution:
          3 5 1
                    2 4    1 4    1 2
       a) 1 2 4 = 3     −5     +1     = 3(−8) − 5(−14) + (−5) = 41
                    3 2    4 2    4 3
          4 3 2
          3 5 1
                    1 4    3 1    3 1
       b) 1 2 4 =−5     +2     −3     =−5(−14) + 2(2) − 3(11) =41
                    4 2    4 2    1 4
          4 3 2
2.   Expanding along which row or column will make the calculation of the
     determinant
 7 23 1
 0 −7 0 easier. Write down the value of the determinant.
13 −15 2
                           a c      a c
  1. Differentiate between      and
                           b d      b d
                                                 1 1 1 
                                                       
  2. Calculate the determinant of the matrix M = 1 2 3  by:
                                                 1 3 2 
                                                       
          5 0                 4k   −20k 
  a) X =              b) X =             , where k is a constant real
          −3 9                −k    5k 
    number.
                                                  
                                   12        0 351
                     1 3                        
             a) A =      b) B =  3         0 16 
                     2 6         5              
                                  −          0 −9 
                                   7              
                                         −1 2           3 −5 
       2.     Consider the matrices A =        and B =        .;
                                         3 −5           −1 2 
                            −1 2     3 −5
             b) Calculate        and
                            3 −5     −1 2
CONTENT SUMMARY
A square matrix can be changed into simpler for before calculating its
determinant through properties including the following:
                                                  m
  1. If all the entries of a row or column of a square matrix are zeros, then the
      determinant of the matrix is zero.
  2.        If all the entries of a row (or column) of a square matrix are multiplied
            by a real number k , then the determinant of the matrix is multiplied by
                        a a ' a ''       a a ' a ''
                       kb kb ' kb '' = k b b ' b ''
            k .Thus, c c ' c ''          c c ' c ''
Example 1.3.2.
                                                    3 5 4
                                                          
  1.   Calculate the determinant of the matrix A =  1 4 2  , then, wi hout
       further calculation, find:                   2 3 2
                                                                     t
           3 4 5
           1 2 4
           2 2 3
       a)
            8 5 4
            5 4 2
            5 3 2
       b)
Solution:
3 5 4
         4 2    1 2    1 4
1 4 2 =3     −5     +4     =3(8 − 6) − 5(2 − 4) + 4(3 − 8) =−4
         3 2    2 2    2 3
2 3 2
           3 4 5
Therefore, 1 2 4 = −(−4) = 4
           2 2 3
                8 5 4 3 5 4
                5 4 2 = 1 4 2 = −4
                5 3 2 2 3 2
		
2. Evaluate
                26 31 −1
                53 9 42
                0 0 0
		
Solution:
                26 31 −1
                53 9 42 =0, since row 3 consists of zeros only.
                0 0 0
		
                   a c ka kc
            a) k      =
                   b d kb kd
                  a c   ka kc 
            b) k      =       
                  b d   kb kd 
                                                                 1 1 1
     2.     Apply the indicated transformations to evaluate 1 2 1
                                                            1 1 2
            a) R2 → R2 − R1 and R3 → R3 − R1
            b) Write down the value of the determinant
                      1 1 1 
                            
 Given the matrix A = 1 2 1 
                      1 1 2 
                            
 a) Use augmented matrix and the following elementary row operations
i) R2 → R2 − R1 and R3 → R3 − R1
ii) R1 → R1 − R3
iii) R1 → R1 − R2
                                       1
       iv) Write down matrix X =           Adj ( A)
                                     det A
CONTENT SUMMARY
Let A be a square matrix of order 2 × 2 or 3 × 3 .Then the inverse of A . Can also
be calculated through the following four steps:
  1.    Find the determinant of A , that is det A ;
                                                     1
  4.    The inverse of matrix A is         A−1 =         Adj ( A) , where matrix A is
        regular, or invertible.                    det A
                      a c                                 −1   1  d −c 
In particular, if A =      is an invertible matrix, then A =             
                      b d                                    det A  −b a 
Example 1.4.1.
Find the inverse of each of the following matrices:
Solution:
     1  8 −3 
A−1 =        
     2  −2 1 
        3 5 1
                  2 4    1 4 1 2
det B = 1 2 4 = 3     −5    +    = 41
                  3 2    4 2 4 3
        4 3 2
Matrix of cofactors:
  2     4       1 4     1 2
 +          −         +      
  3     2       4 2     4 3
  5                             −8 14 −5 
         1       3 1     3 5             
 −
C=           +         −      =  −7 2 11 
  3     2       4 2     4 3            
                              18 −11 1 
 + 5    1
             −
                 3 1
                       +
                         3 5
  2     4       1 4     1 2 
 
                           −8 −7 18 
                            T       
Adjoint matrix: Adj (B)
                     = C=  14 2 −11
                           −5 11 1 
                                    
                     −8 −7 18 
                 1 
                  −1           
      =
The inverse is B     14 2 −11
                 41           
                     −5 11 1 
                  5 0 2
                        
          b) M =  3 4 5 
                  2 1 2
                                         3 7 2           2 −2 1 
                                                 
2) Consider of the following matrices A =  1 0 2  and B =  3 1 6 
                                          0 1 1                   
                                                           −1 1 1 
Find:                                                               
   i)     A−1 and B −1
   ii)    (A )
            −1 −1
           (10A)
                    −1
   iii)
                     Material(tiles) Labor(hours)
Floor tiles          4               3
Wall tiles           11              1
Given that the total cost for floor tiles is 53 (thousand) FRW and the total cost
for wall tiles is 37(thousand) FRW, find the material cost and the labor cost
by answering the following questions:
   a) Label x the material cost and y the labor cost, and then model the
      problem by simultaneous linear equations in x and y
    CONTENT SUMMARY
    The two simultaneous linear equations in two unknowns, x and y ,
      ax + by =    c                                          a b  x   c 
                       can be arranged, using matrices as,             =  
     a ' x + b ' y =c'                                        a ' b '  y   c '
        a b           x         c
Let A =
=                 , X   and B =   .
         a ' b '      y          c '
Then A. X = B .Multiplying both sides by A−1 , we have A−1 .( A. X ) = A−1 .B ,using
the associative property, the inverse property , and the identity property of
multiplication of matrices , we have:
In the same way, the three simultaneous linear equations in three unknowns,
                  ax + by + cz =          d
                 
     x , y and z  a ' x + b ' y + c ' z =   d ' can be arranged, using matrices as, 		
                 a '' x + b '' y + c '' x = d ''
                 
                     a b c   x  d 
                                       
                     a ' b ' c '  . y  =  d ' 
                     a '' b '' c ''   z   d '' 
                                       
                X = A−1.B
       Example 1.4.2.
       Use matrices to solve the following simultaneous equations:
                   2 x − 3 y =
                              −8
                a) 
                    5x + y = −3
                   2 x − y + z =−1
                   
                b)  x + 3 y − z =
                                 13
                    x+ y+z =     3
                   
       Solution:
             a) The system of the simultaneous equations can be expressed as
                    2 −3   x   −8 
                           =  
                    5 1   y   −3 
          2 −3       x         −8 
   Let A =
   =             ; X   and B =   . The system becomes A. X = B ;
         5 1         y         −3 
X = A−1.B ;
                 2 −3                           1  1 3
       det A=        = 2(1) − 5(−3)= 17 ; A−1 =         ;
                 5 1                           17  −5 2 
            1  1 3   −8  1  −17   −1
  =X              =   =             , that is x = −1 and y = 2
           17  −5 2   −3  17  34   2 
          2 −1 1   x   −1
                    
          1 3 −1  y  = 13 
         1 1 1  z   3 
                    
       2 −1 1              x           −1
                                       
    A  1 3 −1 ; =
Let =             X          y  and B =  13  .
      1 1 1               z           3
                                       
      2 −1 1
              −1 1   2 1  2 −1
det
=   A 1 3= −1      −     + = 8 (Expansion along the third
row). 1 1 1   3 −1 1 −1 1 3
Matrix of cofactors:
  3 −1              1 −1       1 3 
  +             −            +     
  1 1               1 1        1 1 
  −1 1                                4 −2 −2 
                      2 1      2 −1           
  −
C=                +          −      =   2 1 −3 
     1 1             1 1      1 1  
                                                
                                     −2 3 7 
  −1 1              2 1       2 −1 
  + 3 −1        −
                     1 −1
                             +
                               1 3 
 
                                  4 2 −2 
                                          
The adjoint matrix is Adj ( A) =  −2 1 3  ,
                                  −2 −3 7 
                                          
                       4 2 −2 
                 1   1              
               A−=     −2 1 3 
                     8              
 he inverse is         −2 −3 7  ;
T   4 2 −2   −1    16   2 
  1           1          
X = −2 1 3   13  =  24  =
                               3
  8              8       
    −2 −3 7   3    −16   −2  , that is x = 2 , y = 3 and z = −2
            4 x + 7 y =
                       34
         a) 
             3x − 2 y =
                       11
             x − 2 y + z =−2
            
         b)  3 x + y − 2 z =7
             x + 3y − z =  2
            
                                              ax + by =    c(1)
  Consider the simultaneous linear equations 
                                             a ' x + b ' y =c '(2)
CONTENT SUMMARY
To solve the two simultaneous linear equations in two unknowns x and y ,
Cramer’s rule requires to go through the following steps:
                                    ax + by = c
  1.   Arrange the equations to get               and calculate the principal
                                    a′x + b′y =c′
                         a b
       determinant = D       = ab′ − a′b
                         a ′ b′
If D = 0 , then the system has no solution or infinitely many solutions; the
system is not a Cramer’s system.
                                                             c b
  2.   Write       down       and      calculate:      D=
                                                        x       = cb′ − c′b           and
                                                             c′ b′
               a c
       D=
        y          = ac′ − a′c
               a ′ c′
                                            Dx         Dy
  3.   Write down and calculate x =            and y =    ; the solution set of the
                                            D          D
                                      D Dy  
       simultaneous equations is S =  x ,   
                                       D D  
                                 ax + by + cz =    d
                                 
         unknowns, x , y and z , a′x + b′y + c′z =   d′
                                 a′′x + b′′y + c′′x = d ′′
                                 
                                          a b c
         The principal determinant is D = a′ b′ c′
                                          a′′ b′′ c′′
         If D = 0 , then the system is not a Cramer’s system, it may have zero solution
         or infinitely many solutions.
                                            D Dy Dz  
         solution; the solution set is S =  x ,  ,     , where
                                             D D D  
     d b c           a d c          a b d
=Dx d= ′ b′ c′ , Dy a=′ d ′ c′ , Dz a′ b′ d ′
    d ′′ b′′ c′′    a′′ d ′′ c′′    a′′ b′′ d ′′
Example 1.4.3.
                                                                3 x + 2 y − 2 z = 4
                                                                
         Use Cramer’s rule to solve the simultaneous equations:  x + 3 y + z = 7
                                                                2 x + y − z =  11
                                                                
                       3 2 −2
                                 3 1     1 1     1 3
         Solution: D = 1 3 1 = 3      −2      −2     = 4
                                 1 −1    2 −1    2 1
                       2 1 −1
   Dx 72
x
=   = = 18
   D   4
   Dy −36
y=   =    = −9
   D    4
   Dz 64
z =
=      = 16
   D   4
              4 x + 7 y =
                         34
           a) 
               3x − 2 y =
                         11
               x − 2 y + z =−2
              
           b)  3 x + y − 2 z =7
               x + 3y − z =  2
              
             −3 4 2 
     a) A =                      B
                                 b)=     (4   −1)
             1 5 −1
2. Given that matrices A and B are equal, find the values of the letters:
             3m + 2 2           m − 3 2
     a) A =            and B =         
             2k + 1 1           5     6
             x −1    −3              6x + 5 2 
     b) A =                 and B =           
             x + 4z 3y + 4           1      1
           −2 3   1 4 
     a) 3        − 2  
           1 2   −1 −1
                   1 3
         −1 3 1       
     b)           1 −1
         2 4 −1  −1 2 
                        
                     2 0 1             1 0 1 
                                             
5. a) Given that A =  3 0 0  , and B = 1 2 1  , find the products
                     5 1 1             1 1 0 
                                             
		 A.B and B. A
                                           2 0 1
                  −1 4                          
          a) A =                  b) M =  3 0 0 
                  2 3                    5 1 1
                                                  
                                       1   2  x2         1 4 1
                                                     T         
     7. a) The transpose of matrix M =  4  1  0  is M =  2 1 1  .
                                       1 x +3 8          4 0 8
                                                               
                       a c            a' c'
        b) Use A =          and B =            to derive the formulas for :
                       b d             b ' d '
     i) ( AT )T : the transpose of the transpose of a matrix
                                  3 2 1
                12 6
          a) a)               b) 0 2 −5
                 5 4
                                 −2 1 4
                                   2 0 0
     9.   a) In the calculation of 1 2 0 , along which row or column is it
                                   3 5 6
                                                        2 1 2
 ii) Name the transformation applied on the determinant 1 2 0 to
                                                        3 5 6
                           2 1 7
    obtain the determinant 1 2 4 .
                           3 5 17
                             2 1 7
iii) write down the value of 1 2 4
10.                          3 5 17
   a) Use inverse matrix to solve the simultaneous linear equations:
        x+ y−z =     0
       
        x + 2 y + 3z =
                      14
       2 x + y + 4 z =
                      16
       
   b) Use Cramer’s rule to solve the simultaneous equations:
       x+ y+z =     6
      
       2x + y − z =  1
      3 x + 2 y + z =
                     10
      
Introductory activity
                     y 3 x − 2 (1), and =
  Consider functions =                    y x 2 + 1 (2)
        a) complete the following table for each of the two functions:
      x1    x2        ∆x = x2 − x1    y1           y2   ∆y = y2 − y1   ∆y
                                                                       ∆x
      1     2
      1     1.5
      1     1.1
      …     …        …                …            …    …              …
                                      ∆y
            b) How is the quantity       called?
                                      ∆x
                           ∆y
            c) Compare        for function (1) and for function (2)
                           ∆x
                                                   ∆y
            d) Predict a formula for lim
                                           ∆x →0   ∆x
CONTENT SUMMARY
                                                                  ∆y f ( x2 ) − f ( x1 )
x assumes values from x1 to x2 , is the quantity defined by          =
                                                                  ∆x      x2 − x1
Example 2.1.1.
The cost (in thousand FRW) of producing a certain commodity is modeled by
the function C ( x=
                  ) 40 + x . Find the average rate of change of the cost when
the production level changes from x = 64 to x = 100 .
Solution:
                        ∆C  2  1
change in the cost is    = =
                        ∆x 36 18
                                                                        1
          2.                                                     ) x 2 − as x
                Find the average rate of change of function f ( x=
                                                                        x
                assumes values from 2 to 4 .
∆x = x1 − x0 etc ∆x → ...
∆y = y1 − y0 etc /////////////
     ∆y                                                     etc     ∆y
                                                                       → ...
     ∆x                                                             ∆x
                                                       ∆y
                                                          →b
               b) How is the statement “If ∆x → a then ∆x    ” written in terms
                  of limits?
                                              ∆y
               c) Find, in terms of x , lim      for y = 3 x 2 at any value of x .
                                        ∆x →0 ∆x
                    f ( x0 + ∆x) − f ( x0 )
number, lim                                 , provided it exists.
            ∆x →0             ∆x
This number is called the derivative of function y = f ( x) at x0 ; it is denoted by
f '( xo ) . More generally, the derivative of function y = f ( x) is denoted by f '( x)
       dy
or y ' or
       dx
Example 2.1.2.
                                                1
                     =
Differentiate function    ( x)
                       y f=                       from first principles.
                                                x
Solution:
                              f ( x + ∆x) − f ( x)
        f '( x) = lim
                      ∆x →0            ∆x
                 1      1
                      −
        = lim x + ∆x x
          ∆x →0    ∆x
                       −1
        = lim
            ∆x →0 x ( x + ∆x )
             −1
        =
             x2
                                         1 4
  f ( x) = 3x 2 − 2 x + x + 1 ; g ( x) =   x + 5 x − 7 and h( x)= 4 x −3 + 7 x − 5
                                         3
           a) Which of the three functions is a polynomial function?
           b) Describe the quantities and the operations involved in a
              polynomial
           c) Obtain, from first principles, the derivative of each of the following:
     i) A constant function C
        d
           (a0 + a1 x + a2 x 2 + ... + an x n ) = a1 + 2a2 x + ... + nan x n −1
        dx
Example 2.2.1.
Find the derivative of each of the following polynomials:
                      3     2
       a) f ( x) = 2 x − 5 x + 11x − 3
                     1      3
       b) g ( x) = − x 3 + x 2 − 25
                     3      4
Solution:
                      2
       a) f '(x) = 6 x − 10 x + 11
       b) g '( x) =       3
                   − x2 + x
                          2
                        dz     dP
   Find the derivative     or     of:
                        dt     dQ
                      3     5
          a) z =2 − 8t + 3t
          b) P = 5Q 4 + 3Q − 7
                           d
           ii) Calculate      (uv)
                           dx
                      du     dv
         b) i) Find      and
                      dx     dx
                              du    dv
           ii) Calculate: v      +u
                              dx    dx
                        d             du    dv
         c) Compare        (uv) and v    +u
                        dx            dx    dx
CONTENT SUMMARY
To find the derivative of the product of two functions u and v of independent
variable x , one can proceed as follows: Either expand the product, and then
find the derivative of the expansion, Or, calculate the derivative of each factor,
                                   du       dv
and then obtain the products v        and u , and finally, consider the sum
                                   dx       dx
   du     dv          d          d       d
v      + u , that is, =  (uv) v (u ) + u (v)
   dx     dx          dx        dx       dx
It is the matter of choosing the more convenient method.
                 1     3
                − x 3 ( x 2 − 25)
      b) g ( x) =
                 3     4
Solution:
                                        d                        d
   u 2 x 3 − 5 x 2 and=
Let=                  v 11x − 3 . Then,     u ) 6 x 2 − 10 x and
                                           (=                       (v) = 11
                                        dx                       dx
Substituting these values in the product rule formula,
 d
    (uv) = (11x − 3)(6 x 2 − 10 x) + 11(2 x3 − 5 x 2 )
 dx
		 =88 x3 − 183 x 2 + 30 x
         1            3 2             d                d        3
Let u = − x 3 and
                =  v    x − 25 , Then (u ) = − x 2 and    (v ) = x
         3            4               dx               dx       2
Substituting these values in the product rule formula,
d         3                      1     3
   (uv
    =  ) ( x 2 − 25)(− x 2 ) + (− x3 )( x)
dx        4                      3     2
        5
       − x 4 + 25 x 2
       =
        4
                          dz    dP
   Find the derivative       or    of:
                          dt    dQ
               (2 8t 3 )(4 + 3t 5 )
          a) z =−
        b) P 5Q 4 (3Q − 7)
        =
     y (2 x + 1)3
  Let=
                                               dy
      a) Expand y and then calculate
                                               dx
                   d
      b) i)Find       (2 x + 1)
                   dx
                                d
        ii) Calculate 3(2 x + 1) 2(2 x + 1)
                               dx
        iii) Describe how to obtain 3(2 x + 1) 2 from (2 x + 1)3
                                        d n
      c) Predict a rule for finding        (u ) , where u is function of variable x
                                        dx
CONTENT SUMMARY
         n
Let y = u , where u is function of variable x and n a rational number. It can be
             d n              du
               (u ) = nu n −1
shown that dx                 dx
 d n                du
     (u ) = nu n −1
 dx                 dx
It is the matter of choosing the more convenient method.
Example 2.2.3
Find the derivative of each of the following powers:
        a) f ( x)= (2 x3 + 3 x 2 − 5 x + 1)6
            ( x)
        b) g=         3x + 2
                              3
                       =
                           2 3x + 2
       c) P 5Q 4 (3Q − 7)
       =
    y (2 x + 1)3
 Let=
                                          dy
       a) Use the power rule to calculate
                                          dx
       b) i) Determine two functions u ( x) and v(u) such that                       y = vu ,
             that is y = v[u ( x)]
                       du     dv dy
            ii) Find      and   =
                       dx     du du
                                   dy du
            iii)   Calculate         .
                                   du dx
                                       d n
         d) Predict a rule for finding    (u ) , where u is function of variable
            x                          dx
CONTENT SUMMARY
Consider the following diagram:
Example 2.2.4.
Use the chain rule to find the derivative of:
         y (4 x + 5)6
      a) =
b)=y 3x + 2
Solution:
                            du
       u 4 x + 5 .Then
a) Let =                       = 4 , the function becomes y = u 6 . We have
                            dx
dy
   u 5 6(4 x + 5)5 ,
= 6=
du
                         dy dy du
From the chain rule,       = .    =6(4 x + 5)5 (4) =24(4 x + 5)5
                         dx du dx
                     dy dy du    1             3
                 = =
From the chain rule,      .         =   (3)
                     dx du dx 2 3 x + 2     2 3x + 2
                              3x 2
  Consider the function y =
                             2x +1
        a) i) Express y in the form y = u.v −1 , stating the value of u and the
           value of v
ii) Use the product rule to find the derivative of y = u.v −1 and, express
                                                                                N ( x)
        your answer with positive exponents, in the form of a fraction
                                                                                D( x)
                                     d                     d
                         (3 x + 1)      (3 x 2 ) − (3 x 2 ) (3 x + 1)
        b) Calculate                 dx                    dx
                                           (3 x + 1) 2
                                             du     dv
                                         v      −u
                   d u                       dx     dx
        c) Compare   ( ) and                      2
                   dx v                         v
                                     d u
        d) Predict a rule for finding ( ) , where u and v are functions of
           variable x                dx v
                          du    dv
                      v      −u                     '
     d u                dx    dx , that is,    u  u 'v − v 'u
Then     =                                      =
     dx  v                 v2                 v       v2
Example 2.2.5.
Use the quotient rule to find the derivative of:
         x 2 − 3x + 2
a) y =
             2x + 1
         2− x
b) y =
            x
Solution:
Let u = x 2 − 3 x + 2 and =
                          v 2x +1 .
     du            dv
Then = 2 x − 3 and    = 2.
     dx            dx
            dy (2 x + 1)(2 x − 3) − 2(x 2 − 3 x + 2)
We have        =
            dx               (2 x + 1) 2
                    2 x 2 + 3x − 7
                =
                      (2 x + 1) 2
Let u= 2 − x and v = x .
         du          dv   1
Then        = −1 and    =   .
         dx          dx 2 x
                 2− x
            − x−
 =
We have:
         dy
            =    2 x −x − 2
         dx    x      2x x
                     5x − 6
           a) y =
                     7x − 4
                      3x 4
           b) y =
                     2 + 5x
Consider function f ( x) = ln x
      a) Using a calculator, complete the following table:
 x0
             1                       2                         3                        …
      ∆x
              ∆y                      ∆y                       ∆y
              ∆x                      ∆x                       ∆x
 0.1            ln(1 + 0.1) − ln1       ln(2 + 0.1) − ln 2       ln(3 + 0.1) − ln 3
              =                       =                        =
                       0.1                     0.1                      0.1
              =                       =                        =
              ∆y                      ∆y                       ∆y
              ∆x                      ∆x                       ∆x
 0.01           ln(1 + 0.01) − ln1      ln(2 + 0.01) − ln 2      ln(3 + 0.01) − ln 3
              =                       =                        =
                       0.01                    0.01                     0.01
              =                       =                        =
             ∆y                       ∆y                       ∆y
             ∆x                       ∆x                       ∆x
 0.001         ln(1 + 0.001) − ln1      ln(2 + 0.001) − ln 2     ln(3 + 0.001) − ln 3
             =                        =                        =
                      0.001                    0.001                    0.001
             =                        =                        =
 …
b) Use the table above to write the values of f '(1); f '(2); f '(3);... f '( x)
                                                                                  1
The derivative of function=   ( x) ln x is
                           y f=                                f '( x) (ln
                                                               =        =  x) '     , that is
                                                                                  x
 d            1
    ( ln x ) = , where x > 0 . More generally, if y = ln[u ( x)] , where u ( x) is function
 dx           x
                                              d               1 du                       u'
of variable x , then from the chain rule, [ lnu(x) ] = . , that is [ ln u ( x) ] ' = .
                                              dx              u dx                        u
                                1                               1               1
In the same way, if log a x =       .ln x , then=( log a x ) ' =    (ln x) '        , where
                               ln a                            ln a          x ln a
a > 0 and a ≠ 1 , and x > 0 .
More generally,
                                                                                             u'
If y = log a u ( x) , where u ( x) is function of variable x , then [ log a u ( x) ] ' =
                                                                                           u ln a
Example 2.2.6.
Find the derivative of:
       a)=y ln( x 2 + 3)
       b)=y log(1 − x 2 )
Solution:
                   2x
       a) y ' =    2
                  x +3
                       −2 x
       b) y ' =
                  (1 − x 2 ) ln10
i) (a x ) ' = ...
i) (a u ) '
CONTENT SUMMARY
                                                                            d x
                          =
  he derivative of function       ( x) e x is f =
                               y f=                     x
                                                '( x) (e= ) ' e x , that is dx
                                                                               e = ex       ( )
T                          u ( x)
. More generally, if y = e , where u ( x) is function of variable x , then from
                  d u ( x)             du
the chain rule,      e  = eu ( x )    , that is eu ( x )  ' = u ' eu . In the same way, if a x ,
                 dx                    dx
      ( )
then a x ' = a x ln a , where a > 0 and a ≠ 1 , and x > 0 . More generally, If y = a u ( x )
Example 2.2.7.
Find the derivative of:
                         2
                             +3
          a) y = e x
                                  2
          b) y = 101− x
 a) Write down the gradient of the straight line through points A and B
 b) As point B moves along the curve towards point A :
      i) The change in x , that is ∆x approaches which value?
      ii) The gradient of the straight line through points A and B approaches
          which value? (Express your answer in terms of the derivative of
         function f
      iii) How becomes the position of the line through points A and B with
         respect to the graph of function y = f ( x)
 c) Write down the equation of the straight line through points A and T .
Example 2.3.1.
Solution:
       a)        f ( x0 ) =      −2(10) 2 + 80(10) =
                          f (10) =                 600 ;                  f '( x) =−4 x + 80 ;
             f '(10) =
                     −4(10) + 80 =
                                 40 ;
The equation of the tangent is: y − 600 = 40( x − 10) ; or, equivalently,=
                                                                         y 40 x + 200
       b) f ( x0 )=   f (19)= 200 − 40 ln(19 + 1)= 200 − 40 ln 20 ≈ 80.17 ;
        −40               −40
f '( x) =    ; f '(19) =        = −2 ;
        x +1             19 + 1
The equation of the tangent is: y − 80.17 =
                                          −2( x − 19) ; or, equivalently,
y= −2 x + 98.17
                            1
            b)   f (x) = 10- x ; x0 (Leave your answer in terms of surds)
                            2
                                                                f ( x) f ( x0 ) 0
  Let numerical functions f ( x) and g ( x) be such that lim = =
                                                         x → x0 g ( x ) g ( x0 ) 0
     ∞
  or
     ∞
        a) Determine the equations of the tangents to the graphs of functions
              f ( x) and g ( x) at x0 , in the form =
                                                    y ax + b and =
                                                                 y cx + d
                                       f ( x)
           b) Calculate lim                   , substituting f ( x) , and g ( x) , respectively by
                              x → x0   g ( x)
              ax + b and cx + d , obtained from (a) and simplify.
                                       f ( x)
           c) Express lim                     in terms of a limit involving f '(x 0 ) and g '( x0 )
                             x → x0    g ( x)
CONTENT SUMMARY
                                                           f ( x) f ( x0 ) 0
                                                       lim = =
If num ric l functions f ( x) and g ( x) are such that     g ( x) g ( x0 ) 0 or
                                                                       x → x0
 ∞     e a
 ∞ , then to remove the indetermination, we proceed as follows, through
Hospital’s rule:
  1.     Differentiate separately, the numerator and the denominator, to get
          f '(x) and g '( x) ;
                              f '( x) f '( x0 )
  2.     Calculate lim               =
                    x → x0    g '( x) g '( x0 )
                    f ( x) f '( x0 )
  3.     Then lim           =
             x → x0 g ( x )   g '( x0 )
Note that: - the process can be repeated if necessary;
                                                                                     0 ∞
        – Hospital’s rule is used only if we have indetermination                     or
                                                                                     0 ∞
                                            '
               f ( x)           f ( x) 
        lim            ≠ lim           
        x → x0 g ( x )   x → x0 g ( x )
                                       
       0 ∞
         or
       0 ∞
Example 2.3.2.
Evaluate the following limits:
             1 − ln x
        lim
         x →e x − e
            e3 x − 1
        lim
       x →0    x2
Solution:
            1 − ln x 1 − ln e 1 − 1 0
       a) lim   =         = =         : indeterminate case
              x−e
            x →e        e−e   e−e 0
Applying Hospital’s rule,
                                              1
                                            −
             1 − ln x      (1 − ln x) '                1   1
We have: lim          lim
                      =                 lim x =
                                        =       − lim =  −
         x →e x − e   x → e ( x − e) '  x →e 1    x →e x   e
               e3 x − 3 x − 1 e3(0) − 3(0) − 1 1 − 1 0
       b) lim            =                = =          : indeterminate case
          x →0       x2              02         02   0
                          x −1
           a) lim
               x →1   ( x − 1)e x
                      2 x−2
           b) lim e         −1
               x →1 ln(5 x − 4)
a) f ( x) = 3x 2
                    −5
          b) f ( x) =
                     x
     3.   Find the derivative of:
             y 30 x − 0.5 x 2
          a) =
                  1
          b) y =+ 2 x − 3
                  x
     4.   Differentiate, with respect to t :
          a) s 3t 4 (2t − 5)
          =
          b) s =(t 7 − 4)(t 5 + 11)
                  dy
     5.   Find       ,using the chain rule, if:
                  dx
          a) y=        x2 − x + 2
          b)=y (3 x 4 + 7)6
                 5P 2 − 9 P + 8
     a) Q =
                     P2 + 1
                 6P − 7
     b) Q =
                 8P − 5
7.   Differentiate, with respect to x :
                (3 x + 4)3
     a) y =
                  5x −1
                             2
             2x +1 
     b) y =         
             3x − 5 
8.   Find the derivative of;
a) y = 21−3 x
b) y = 3ln 2 (5 x )
a) f ( x) = −2e3 x ; x0 = 0
                 x
     b)     f ( x) =; x0 = e 2
               ln x
10. Calculate the following limits:
                 e 2 x − ln( x + e)
     a) lim
            x →0          x
                  ex
     b) lim
           x →0 e x + 1
Introductory activity
                   dC
Marginal Cost =
                   dq
Example 3.1.1.
A radio manufacturer produces x sets per week at a total cost of
     1 2
y
=      x + 3 x + 100 FRW.
    25
Find the marginal cost for x = 30
Solution:
d 1 2               2
  ( x + 3 x + 100)=    x+3
dx 25               25
                                                  2
For, x = 30 , the value of the marginal cost is      (30) + 3 =5.4 per unit.
                                                  25
Application activity3.1.1
                           1
          Q = 40 P + 3P 2 − P 3 , where Q is the total output and P is the
                           3
          number of units of input. Find the value of the marginal product for
          P = 10
                                            P 100 − Q .
  A firm has the following demand function: =
  Find: a) in terms of Q , the total revenue function
           b) The instantaneous rate of change of the total revenue when Q = 11 .
CONTENT SUMMARY
Solution:
dR                         −3Q + 32
= Q ' 16 − Q + Q( 16 − Q=
                        )'          ,
dQ                         2 16 − Q
          dR −3(7) + 32 11
     = 7,
For Q=       =
          dQ 2 16 − 7    6
CONTENT SUMMARY
                                                           d  dy 
                                                               >0
If function y = f ( x) i such that dy = 0 at x0        and dx  dx  at x0 , then
                        s            dx
the function y = f ( x) has a local minimum at x0 ; the minimum value of the
function is f ( x0 ) .
                                                                               dy
Function y = f ( x) is said to be increasing for the values of x such that        > 0,
                                                                               dx
                                                   dy
and decreasing for the values of x such that          < 0,
                                                   dx
Example 3.2.1.
                                1 3 9 2
Given the total cost function P = Q − Q + 14Q + 22 , find the value of Q for
                                3       2
which the total cost is minimum, and find the minimum total cost
Solution:
dP
   = Q 2 − 9Q + 14
dQ
dP
   = 0 if and only if Q 2 − 9Q + 14 =
                                    0 ; solving this quadratic equation, we get
dQ
Q1 = 2 and Q2 = 7 ;
Therefore, the minimum value of the total cost occurs for Q = 7 , and the
                              1 3 9 2
corresponding total cost is     (7) − (7) + 14(7) + 22 =
                                                       13.83
                              3      2
   Find the value of Q for which the total cost is minimum, and find the
   minimum total cost in each of the following cases:
           a) P= ln(Q 2 − 8Q + 20)
                    1 3 17 2
           b) P =     Q − Q + 60Q + 27
                    3    2
CONTENT SUMMARY
Given the total cost function, P = 5e −0.2Q , find the value of Q for which the
total revenue is Given the demand function maximum, and find the maximum
revenue.
Solution:
                                                  dR
The total revenue is R(Q) = 5Qe −0.2Q . Then,       = (5 − Q)e −0.2Q ;
                                                  dQ
dR                             d
   = 0 if and only if Q = 5 ;    [(5 − Q)e −0.2Q ] = (−2 + 0.2Q)e −0.2Q ;
dQ                            dQ
                              1
its value at Q = 5 is −e −1 =− < 0 ;
                              e
Therefore, the maximum value of the total revenue occurs for Q = 5 and the
                             25
maximum value is R(5)
                   =         = 9.1
                              e
                        dQd P                                   dQ P
         b) Calculate       .  , for P = 10 . Predict a name for d .
                         dP Qd                                   dP Qd
If ε d < 0 , then Qd and P are such that the increase in P implies the decrease in
Qd
Example 3.3.1.
Solution:
dQd
    =−5 − 2 P ; Elasticity of the demand:
 dP
          dQd P                        10
      εd =    .  =[−5 − 2(10)]                      =−0.5
           dP Qd               650 − 5(10) − (10) 2
     Find the price elasticity of the demand if the quantity demanded Qd and
     the price P are related by:
                 100
         a) Qd ln=
         =
                  P2
                     ;P 4
                      20
          b) Qd
          =          =    ;P 3
                     P +1
  A beverage company estimates that the amount Qs of soft drink supplied per
  month and the quantity C bought by customers are related by a function
  Qs = f (C ) .It is observed that C increases, for a particular value of C .
CONTENT SUMMARY
                                                      dQs P
The price elasticity of supply is defined by, ε s =      . ,
                                                      dP Qs
Where Qs : quantity supplied, and P : amount received from consumers.
Price elasticity of supply indicates how producers respond to the change in the
amount they receive from the consumers
Example 3.3.2.
Solution:
               1   2 dQs 1                  1       2
We have: Q
         =s      P+ ;   = ; For P = 3 , Qs=   (3) + = 1
               5   5 dP 5                   5       5
                                        dQs P 1 3 3
                                 εs
The price elasticity of supply is=         .= =.    ,
                                        dP Qs 5 1 5
 Find the price elasticity of the supply if the quantity supplied Qs and the
 price P are related by:
       a) P= 3 + 4Qs ; P = 11
             Pe −0.2 P ; P 4
      b) Qs 5=
      =
     1.   Find the price and the quantity that will maximize the total revenue,
          given the demand function: P = 12.5e −0.005Q
                                                                            4x
     2.   Find, the coordinates of the minimum point of function, f ( x) =
          , and confirm that it is a minimum.                              3ln x
     5.                                                    Q 150 − 15 P at
          Find the price elasticity of the demand function =
          P = 4 , where P is the price.
Introductory activity
CONTENT SUMMARY
Statistics is the branch of mathematics that deals with data collection,
data organization, summarization, analysis, interpretation, and drawing of
conclusions from numerical facts or data.
Statistics plays a vital role in nearly all businesses and forms the backbone for
all future development strategies. Every business plan starts with extensive
research, which is all compiled into statistics that can influence a final decision.
Statistics helps the businessman to plan production according to the taste of
customers.
Branches of statistics
There are two branches of statistics, namely descriptive, and inferential
statistics.
a. Descriptive statistics
Descriptive statistics deals with describing the population under study. It
consists of the collection, organization, summarization, and presentation of
data in a convenient and usable form.
CONTENT SUMMARY
A variable is a characteristic of interest for each person or object in a population.
A variable is a characteristic under study that takes different values for different
elements. A variable, or random variable, is a characteristic or measurement
that can be determined for each member of a population. For example, if we
want to know the average (mean) amount of money senior five students spend
at Kiziguro secondary school on school supplies that do not include books.
We randomly surveyed 100 first year students at the school. Three of those
students spent 1500Frw, 2000Frw, and 2500Frw, respectively. In this example,
the variable could be the amount of money spent (excluding books) by one
senior five student. Let X = the amount of money spent (excluding books) by
one senior five student attending Kiziguro secondary school. Another example,
if we collect information about income of households, then income is a variable.
These households are expected to have different incomes; also, some of them
may have the same income.
CONTENT SUMMARY
Data are individual items of information that come from a population or sample.
Data is also defined as a set of observations. Data are the values (measurements
or observations) that the variables can assume. They may be numbers, or they
may be words. Datum is a single value. Data may come from a population or
from a sample. Lower case letters like x or y generally are used to represent
data values.
The observations or values that differ significantly from others are called
outliers. Outliers are at the extreme ends of a dataset. Dataset is a collection
of data of any particular study without any manipulation. Information are
facts about something or someone. Most data can be put into the following
categories: qualitative, and quantitative.
Example
You go to the supermarket and purchase three soft drinks (500ml soda, 1ml
milk and 300ml juice) at 5000frw, four different kinds of fruits (apple, mango,
banana and avocado) at 800frw, two different kinds of vegetables (broccoli and
carrots) at 500frw, and two desserts (ice cream and biscuits) at 1000frw.
In this example,
  • Number of soft drinks, different kinds of fruits, different kinds of vegetables,
     and desserts purchased are quantitative discrete data because you count
     them.
  • The prices (5000frw, 800frw, 500frw, and 1000frw) are quantitative
     continuous data.
  • Types of soft drinks, vegetable, fruits, and desserts are qualitative or
     categorical data.
A collection of information which is managed such that it can be updated
and easily accessed is called a database. A software package which can be
used to manipulate, validate and retrieve this database is called a Database
Management System. For example, Airlines use this software package to book
tickets and confirm reservations which are then managed to keep a track of the
schedule.
CONTENT SUMMARY
Variables classified according to how they are categorized or measured. For
example, the data could be organized into specific categories, such as major
field (accounting, finance, etc.), nationality or gender. On the other hand, can the
data values could be ranked, such as grade (A, B, C, D, F) or rating scale (poor,
good, excellent), or they can be classified according to the values obtained from
measurement, such as temperature, heights, or weights.
Therefore, we need to distinguish between them through the measurement
scale used. A scale is a device or an object used to measure or quantifies any
event or another object. In statistics, the variables are defined and categorized
using different levels of measurements. Level of measurement or scale of
measure is a classification that describes the nature of data within the values
assigned to variables (Kirch, 2008).
Nominal scale
A nominal scale is used to name the categories within the variables by providing
no ranking or ordering of values; it simply provides a name for each category
within a variable so that you can track them among your data (Crossman, 2020).
The nominal level of measurement is also known as a categorical measure and
is considered qualitative in nature.
Examples
 • Nominal tracking of gender (male or female)
 • Nominal tracking of travel class (first class, business class and economy
   class).
When the classification takes ranks into consideration, the ordinal level of
measurement is preferred to be used.
Ordinal scale
The ordinal level of measurement classifies data into categories that can be
ordered, however precise differences between the ranks do not exist. Ordinal
scales are used when people want to measure something that is not easily
quantified, like feelings or opinions. Within such a scale the different values
for a variable are progressively ordered, which is what makes the scale useful
and informative. However, it is important to note that the precise differences
between the variable categories are unknowable. Ordinal scales are commonly
used to measure people’s views and opinions on social issues, like quality of the
products, services, or how people are satisfied with something.
Examples
 • if you have a business and you wish to know how people are happy with
   your products or services, you could ask them a question like “How happy
   are you with our products or services?” and provide the following response
   options: “Very happy,” “Somehow happy,” and “Not happy.”
 • To test the quality of the canned product, people can use the rating scale
   either excellent or good or bad.
Interval scale
Unlike nominal and ordinal scales, an interval scale is a numeric one that allows
for ordering of variables and provides a precise, quantifiable understanding of
Example
It is common to measure people’s income as a range like 0Frw-100,000Frw;
100,001Frw-200,000Frw; 200,001Frw-300,000Frw, and so on. These ranges
can be turned into intervals that reflect the increasing level of income, by using
1 to signal the lowest category, 2 the next, then 3, etc.
Ratio scale
The ratio scale is the interval level with additional property that there is also
a natural zero starting point. In this type of scale zero means nothingness.
Another difference lies in that we can attribute some of the quantities to others.
Example
The value of salary for someone is a measurement of type ratio level, where
we can attribute values of wages to each other, as if to say that the person X
receives a salary twice the salary of the person Y. And zero here means that the
person did not receive a salary.
CONTENT SUMMARY
Collecting data on entire population is costly or sometimes impossible.
Therefore, a subset or subgroup of the population can be selected to represent
the entire population. The process of selecting a sample from an entire
population is called sampling. Since the sample selected is representing the
wholepopulationunderstudy,thesamplemusthavethesamecharacteristicsas
the population. There are several ways of selecting sample from the population.
Some of the methods used in selecting samples are simple random sampling,
stratified sampling, cluster sampling, and systematic sampling.
In stratified sampling, the population is divided into groups called strata
and then takes a proportionate number from each stratum. For example, you
can stratify (group) taxpayers by their Ubudehe categories then choose a
proportionate simple random sample from each stratum (Ubudehe category)
to get a stratified random sample. To choose a simple random sample from each
category, number each member of the first category, number each member
of the second category, and do the same for the remaining categories. Then
use simple random sampling to choose proportionate numbers from the first
category and do the same for each of the remaining categories. Those numbers
picked from the first category, picked from the second category, and so on
represent the members who make up the stratified sample.
In cluster sampling, the population is divided the population into clusters
(groups) and then randomly select some of the clusters. All the members from
these clusters are in the cluster sample.
For example, if you randomly sample your costumers by gender (males,
CONTENT SUMMARY
Frequency tables are a great starting place for summarizing and organizing
your data. Once you have a set of data, you may first want to organize it to see
the frequency, or how often each value occurs in the set. Frequency tables can
be used to show either quantitative or categorical data.
Example
Assume that a sample of 50 taxpayers in a district was selected to understand
how taxpayers are satisfied with the taxes they are paying. The responses of
those taxpayers are recorded below where (v) means very high satisfied, (s)
means somewhat satisfied and (n) means not satisfied. v, n, v, n, v, s, n, n, n, n, s,
s, v, n, n, n, s, n, n, s, n, n, n, s, s, s, v, v, s, v, s, v, n, n, n, n, s, v, v, v, v, v, v, v, s, v, v, v, v, v
From the recorded data above, we note that:
 • Eleven of them were not satisfied with the taxes they were paying.
 • Five of them were somewhat satisfied with the taxes were paying.
 • Four of them were very high satisfied with the taxes were paying.
Sum=50
CONTENT SUMMARY
A bar chart or bar graph is a chart or graph that presents numerical data with
rectangular bars with heights or lengths proportional to the values that they
represent. The bars can be plotted vertically or horizontally. A vertical bar chart
is sometimes called a line graph.
To construct a bar graph, we use the following steps:
 • Represent the categories on the horizontal axis (remember to represent all
    categories with equal intervals).
 • Mark the frequencies (or percentages) on the vertical axis.
 • Draw one bar for each category that corresponds to its frequency (or
    percentage) on the vertical axis.
Class S1 S2 S3 S4 S5 S6
Number of learners 45 40 50 60 55 45
Solution
      a) A bar graph showing number of learners per class in a school
Solution
      a) From the bar chart, the highest frequency is 52 and the lowest
         frequency is 23
      b) The bar chart indicating frequency against vehicle color involved in
         total loss collision
Step 1: Draw and label the x and y axes. The x axis is always the horizontal axis,
and the y axis is always the vertical axis.
Step 2: Represent the frequency on the y axis and the class boundaries on the
x axis.
Step 3: Using the frequencies as the heights, draw vertical bars for each class.
See Figure below
Example:
Using the frequency distribution given in Example 2, construct a frequency
polygon
Step 1: Find the midpoints of each class. Recall that midpoints are found by
adding the upper and lower boundaries and dividing by 2:
Step 2: Draw the x and y axes. Label the x axis with the midpoint of each class,
and then use a suitable scale on the y axis for the frequencies.
The frequency polygon and the histogram are two different ways to represent
the same data set. The choice of which one to use is left to the discretion of the
researcher.
c. The cumulative frequency graph or Ogive.
The ogive is a graph that represents the cumulative frequencies for the classes
in a frequency distribution.
Step 1: Find the cumulative frequency for each class.
Step 2: Draw the x and y axes. Label the x axis with the class boundaries. Use
an appropriate scale for the y axis to represent the cumulative frequencies.
Step 4: Starting with the first upper class boundary, 104.5, connect adjacent
points with line segments, as shown in the figure. Then extend the graph to the
first lower class boundary, 99.5, on the x axis.
CONTENT SUMMARY
In most graphs and charts, the independent variable is plotted on the horizontal
axis (the X − axis ) and the dependent variable on the vertical axis (the Y − axis ).
A time series is defined as having the independent variable of time and the
dependent variable as the value of the variable being studied.
A time series graph is a line graph that shows data such as measurements,
sales or frequencies over a given time.
Frequently, “time” is plotted along the x-axis. Such a graph is known as a time-
series graph because on it, changes in a dependent variable (such as GDP: Gross
Domestic Production, inflation rate, or stock prices) can be traced over time.
They can be used to show a pattern or trend in the data and are useful for
making predictions about the future such as weather forecasting or financial
growth.
Example
In a week, a certain company is making a profit of 10000 FRW on the first
day, 15000FRW on the second day, 12000FRW on the third day, 13000FRW
on the fourth day, 9000FRW on the fifth day, 10000FRW on the sixth day, and
9000FRW on the seventh day. In a tabular form, this can be presented as
Day 1 2 3 4 5 6 7
Step 3: To find how many degrees for each pie sector we need, we take a full
circle of 360° and use the formula:
                                Frequency of S × 3600
        Angle for sec tor S =
                                  Total frequency
Since there are 3600 in a circle, the frequency for each class must be converted
into a proportional part of the circle. This conversion is done by using the
formula:
                   f
         Degrees = .3600
                   n
where f is frequency for each class and n is the sum of the frequencies. Hence,
the following conversions are obtained. The degrees should sum to 360.
Step 4: Once all the degrees for creating a pie chart are calculated, draw a circle
(pie chart) using the calculated measurements with the help of a protractor,
and label each section with the name and percentages or degrees.
Example.
  1.   In the summer, a survey was conducted among 400 people about their
       favourite beverages: 2% like cold-drinks, 6% like Iced-tea, 12% like
       Cold-coffee, 24% like Coffee and 56% like Tea.
       a) How many people like tea?
       b) How many more people like coffee than cold coffee?
Solution:
      a) Total number of people = 400
                                             56
      Number of people like tea = 400 ×         = 224
                                            100
                                             24
      a) Number of people like coffee = 400 ×   = 96
                                            100
      Number of people like cold-coffee = 400 ×12 = 48
                                                   6
      a) Number of people like iced-tea = 400 ×       = 24
                                                  100
                                                     2
      Number of people like cold-drinks = 400 ×         =8
                                                    100
                                     24
      Central angle for iced-tea =       × 3600 =21.60
                                     400
                                       8
      Central angle for cold-drinks =     × 3600 =7.20 = 8/400 × 360o = 7.2o
                                      400
      Total central angle = 21.6 + 7.20 = 28.80 = 21.6o + 7.2o = 28.8o.
                                0
Solution
      a) Central      angles    are    calculated      by       using   the     formula:
                     f
           Degrees=    × 3600 and percentages calculated by using the
                     n
                                f
           formula: Percentage=   ×100%
                                n
                      Number
Activity                       Central angle            Percentage
                      of hours
                                    9                   9
Office work           9                × 3600 =
                                              1350         ×100% =
                                                                 37.5% ; 38%
                                    24                  24
                                    1                   1
Exercise              1                × 3600 =
                                              150          ×100% =
                                                                 4.16% ; 4%
                                    24                  24
                                     2                   2
Travelling            2                × 3600 =
                                              300          ×100% =
                                                                 8.33% ; 8%
                                    24                  24
Watching                             3                   3
                      3                × 3600 =
                                              450          ×100% =
                                                                 12.5% ; 13%
shows                               24                  24
                                    7                   7
Sleeping              7                × 3600 =
                                              1050         ×100% =
                                                                 29.16% ; 29%
                                    24                  24
                                     2                   2
Miscellaneous         2                × 3600 =
                                              300          ×100% =
                                                                 8.33% ; 8%
                                    24                  24
Total                 24            3600                100
CONTENT SUMMARY
Once data has been collected, they may be presented or displayed in various
ways including graphs. Such displays make it easier to interpret and compare
the data.
Examples
  1.     The bar graph shows the number of athletes who represented five
         African countries in an international championship.
Solution:
We read the data on the graph:
      a) Total number of athletes are: 18 + 10 + 22 + 6 + 16 = 72 athletes
      b) 6 athletes
      c) 22 athletes
      d) Representation of the given information on the graph on a frequency
         table.
Solution:
      a) To estimate the mode graphically, we identify the bar that represents
         the highest frequency. The mass with the highest frequency is 60 kg.
         It represents the mode.
      b) The highest mass = 67 kg and the lowest mass = 57 kg
Then, The range=highest mass-lowest mass= 67 kg − 57 kg =
                                                        10kg
CONTENT SUMMARY
A measure of central tendency is very important tool that refer to the centre
of a histogram or a frequency distribution curve. There are three measures of
central tendency:
  • Mean
  • Median
  • Mode
Difference between the mean, median and mode
 • Mean is the average of a data set.
 • The median is the middle value in a set of ranked observations. It is also
     defined as the middle value in a list of values arranged in either ascending
     or descending orders.
 • Mode is the most frequently occurring value in a set of values.
∑x i
        1
Or x =
        n
          ∑ xfi
Here, the mean can also be calculated by multiplying each distinct value by its
frequency and then dividing the sum by the total number of data values.
Consider the ranked data x1 , x2 , x3 ,..., xn the formula for calculating the median
for the two cases (even and odd) is given by:
If n is odd, Median = x n +1  , or
                              
                            2 
                                   th
                    n +1 
median is given by        number which is located on this position
                    2 
                           x n  + x n    
                                       +1
                            2         2 
If n is even, Median =                          , or
                                   2
                  1  n   n  
                           th    th
                             803
     • Mean or average is        = 114.714 . This means that mean average of
                              7
       customers in the past week is 115 customers.
The mode is 92 customers because on Monday and Tuesday, 92 customers
were received. To find the median, we need to arrange data as follows: 92, 92,
118, 120, 121, 128, 132. Then, the middle value is 120. Therefore, the median
is 120.
       
  2          .
The value at the fourth position in the ranked data above is 120. Hence,
median is 120.
Example 2
Calculate the mean of the pocket money of some 5 students who get
    2500 FRW , 4000 FRW , 5500 FRW , 7500 FRW and 3000 FRW .
                                =
Divide the sum by the number of students             22500 / 5
                                                     =            4500FRW .
  The heights (at the shoulders) are 600mm, 470mm, 170mm, 430mm, and
  300mm.
        a) Work out the mean height of your dogs.
        b) For each height subtract the mean height and square the difference
           obtained (the squared difference).
        c) Work out the average of those squared differences. What do you
           notice about the average?
CONTENT SUMMARY
The following are measures of variation:
 • Variance
 • Standard deviation
 • Range
 • Mean deviation.
      Σ(x − µ)
                2
σ =
  2
Example1:
The heights (in meters) of six children are 1.42, 1.35, 1.37, 1.50, 1.38 and 1.30.
Calculate the mean height and the variance of the heights.
Solution:
        1
Mean=     (1.42 + 1.35 + 1.37 + 1.50 + 1.38 + 1.30=) 1.39 m
        6
            1
Variance=
            6
              (1.422 + 1.352 + 1.37 2 + 1.502 + 1.382 + 1.302 ) − (1.39 )= 0.00386 m
                                                                         2
Number of
customers
served          20 − 29    30 − 39     40 − 49      50 − 59      60 − 69      70 − 79
lunch
Number
of days in
               6           12          16          14            8            4
the 60-day
period
Find the mean and variance of the number of customers served lunch using this
grouped data.
Solution
To find the mean from grouped data, first we determine the mid-interval values
for all intervals;
                                        ∑ = 60          ∑ = 2850           ∑ = 146635
                2850
        is x
The mean=       = 47.5
                 60
           146635
                  − ( 47.5 ) = 187.67 .
                            2
Variance =
             60
           Σ(x − µ)
                        2
σ =
=   σ      2
                     , where x is individual value, µ is the population
               N
mean, and N is the population size.
 The sample standard deviation is defined as: square root of the average of
 the squared differences from the sample mean.
                        2
                      −
                         
                Σ x − x                                 −
=S      S2
        =                , where x is individual value, x is the sample mean,
                   n −1
 and n is the sample size.
 Example:
 The six runners in a 200 meter race clocked times (in seconds) of 24.2, 23.7,
 25.0, 23.7, 24.0, 24.6.
  Find the mean and standard deviation of these times.
Solution
 σ=
                                                            6
= 0.473 seconds
 Range
 The range for a data set is depends on two values (the smallest and the largest
 values) among all values in such data set. The range is defined as the difference
 between the largest value and the lowest value.
 Mean deviation
 Another measure of variation is called mean deviation; it is the mean of the
 distances between each value and the mean.
 Coefficient of variation
 A coefficient of variation (CV) is one of well-known measures that used to
 compare the variability of two different data sets that have different units of
 measurement.
 Moreover, one disadvantage of the standard deviation that its being a measure
 of absolute variability and not of relative variability.
 The coefficient of variation, denoted by (CV), expresses standard deviation as a
 percentage of the mean and is computed as follows:
                                     σ
 For population data CV=               ×100%
                                     µ
                                S
 For sample data CV=             −
                                     ×100%
                                 x
 Example:
 Two plants C and D of a factory show the following results about the number of
 workers and the wages paid to them.
                                             C                  D
  No. of workers                             5000               6000
  Average monthly wages                      $2500              $2500
  Standard deviation                         9                  10
Solution
To find which plant has greater variability, we need to find the coefficient of variation.
The plant that has a higher coefficient of variation will have greater variability.
Coefficient of variation for plant C:
Using coefficient of variation formula,
          σ
C.V . = ×100%, x ≠ 0
      x
           9
C.V . =        ×100% = 0.36%
          2500
   Rank
            1      2      3      4      5      6      7      8      9      10     11
   order
   Prices
            4500   4600   4800   5200   5400   5600   6300   6400   7700   7700   7800
   (FRW)
CONTENT SUMMARY
Any data set can be divided into four equal parts by using a summary measure
called quartiles.
There are three quartiles that used to divide the data set which is denoted by Qi
for i = 1, 2,3 . The following definition is illustrated the meaning of the quartiles.
Quartiles are three summary measures that divide a ranked data set into four
equal parts. The following are three quartiles:
  • First quartile ( Q1 ) is the middle term among the observations that are less
     than the median.
 • Second quartile ( Q2 ) is the sa e as the median.
                                    m
 • Third quartile ( Q3 ) is the value of the middle term among the observations
     that are greater than the median.
                                                                          th
                                                        3         
end on the rank order. The upper quartile, Q3 takes the  ( n + 1)  position on
                                                        4         
                                                                     th             th
                                                          1         3      
the rank order. For large population, it is enough to use  (n)  and  ( n ) 
positions for the lower and upper quartiles respectively. 4         4      
Example
For the given data set 61, 24, 39, 51, 37, 59, 45. Find the values of the three
quartiles. First, we rank the given data in increasing order. Then we calculate
the three quartiles as follows 24 37 39 45 51 59 61.
  • Median value is 45.
  • Values less than the median are 24 37 and 39.
  • Values greater than the median are 51 59 and 61.
Therefore, the values of the three quartiles are Q1 = 37 , Q2 = 45 Q3 = 59 .
 Figure 1: Three types of skewness (Positive skew, Zero skew or symmetrical distribution, and
                                       Negative skew)
Zero skewness
The symmetrical data has zero skewness as all measures of a central tendency
lies in the middle.   
                                              Mode Median
In summary, for a data set skewed to the left:=    = Mean .
Measure of skewness
It’s one thing to look at two sets of data and determine that one is symmetric
while the other is asymmetric. It’s another to look at two sets of asymmetric
data and say that one is more skewed than the other. It can be very subjective
to determine which is more skewed by simply looking at the graph of the
distribution. This is why there are ways to numerically calculate the measure
of skewness.
One measure of skewness, called Pearson’s first coefficient of skewness, is
to subtract the mean from the mode, and then divide this difference by the
standard deviation of the data.
             Mean − Mode
Skewness =
                  σ
The reason for dividing the difference is so that we have a dimensionless
quantity. This explains why data skewed to the right has positive skewness.
If the data set is skewed to the right, the mean is greater than the mode, and
so subtracting the mode from the mean gives a positive number. A similar
argument explains why data skewed to the left has negative skewness.
Pearson’s second coefficient of skewness is also used to measure the asymmetry
of a data set. For this quantity, we subtract the mode from the median, multiply
this number by three and then divide by the standard deviation.
             3 ( Median − Mode )
Skewness =
                      σ
Examples in real life
Incomes are skewed to the right because even just a few individuals who
earn millions of dollars can greatly affect the mean, and there are no negative
incomes.
Data involving the lifetime of a product, such as a brand of light bulb, are skewed
to the right. Here the smallest that a lifetime can be is zero, and long lasting
light bulbs will impart a positive skewness to the data.
Items 1 2 3 4 5 6 7 8 9 10
Chebyshev’s theorem
Chebyshev’s theorem which applies to any set of measurements (all shapes of
histograms). It states that the proportion of observations
                                                1          that lie within k
                                            1 − 2 , where k > 1 .
standard deviations of the mean is at least     k
When k=2, the Chebyshev’s theorem states that at least three-quarters (75%)
of all observations lie within two standard deviations of the mean.
With k=3, Chebyshev’s theorem states that at least eight-ninths (88.9%) of all
observations lie within three standard deviations of the mean.
CONTENT SUMMARY
With the use of descriptive statistics, we can summarize data related to revenue,
expenses, and profit for companies. For example, a financial analyst who works
for a retail company may calculate the following descriptive statistics during
one business quarter:
  • Mean and median number of daily sales
  • Standard deviation of daily sales
  • Total revenue and total expenses
  • Percentage change in new customers
  • Percentage of products returned by customers.
  • The mean household income.
  • The standard deviation of household incomes.
  • The sum of gross domestic product.
      Annual
                   370    490    580    680    610    640    790    890    980   950
      income
      Household
                   2      4      4      1      3      5      6      4      7     2
      size
      Number of
                   0      0      1      3      2      2      1      1      1     0
      cows
Introductory activity
   Temperature (° C)      14      16     15     16     23     12      21       22
   Ice cream sales        16      18     14     19     43     12      24       26
  If a businessman wants to make future investments based on how ice cream
  sales relate to the temperature,
      i) Which statistical measure and variables will he use?
      ii) Which variable depends on the other?
      iii) In statistics, how do we call a variable which depends on the other?
CONTENT SUMMARY
Bivariate data is data that has been collected in two variables, and each data
point in one variable has a corresponding data point in the other value. Bivariate
dataisobservationontwovariables,whilstunivariatedataisanobservationon
only one variable. We normally collect bivariate data to try and investigate the
relationshipbetweenthetwovariablesandthenusethisrelationshiptoinform
future decisions.
Bivariate statistics deals with the collection, organization, analysis,
interpretation, and drawing of conclusions from bivariate data. Data sets that
contain two variables, such as wage and gender, and consumer price index and
inflationratedataaresaidtobebivariate.Inthecaseofbivariateormultivariate
data sets we are often interested in whether elements that have high values of
one of the variables also have high values of other variables.
In bivariate statistics, we have independent variable and dependent variable.
Dependent variable refers to variable that depends on the other variable (s).
Independent variable refers to variable that affects the other variable (s).
 Figure 5.1: Scatter diagram representing the relationship between secondary school graduate
                                     rate and the poverty.
CONTENT SUMMARY
Covariance is a statistical measure that describes the relationship between a
pair of random variables where change in one variable causes change in another
variable. It takes any value between -infinity to + infinity, where the negative
value represents the negative relationship whereas a positive value represents
the positive relationship. It is used for the linear relationship between variables.
It gives the direction of relationship between variables.
             1 n         −
                                  −
                                                    1 n          − −
cov(x, y)=     ∑ i  i 
                    
             n i =1 
                      x − x   y − y  or cov(x,
                                            =     y)   ∑ i i y.
                                                     n i =1
                                                            x y − x
The covariance of variables x and y is a measure of how these two variables
change together. If the greater values of one variable mainly correspond with the
greater values of the other variable, and the same holds for the smaller values,
i.e., the variables tend to show similar behavior, the covariance is positive. In
the opposite case, when the greater values of one variable mainly correspond
to the smaller values of the other, i.e., the variables tend to show opposite
behavior, the covariance is negative. If covariance is zero, the variables are said
to be uncorrelated, meaning that there is no linear relationship between them.
Correlation
The Pearson’s coefficient of correlation (or product moment coefficient of
correlation or simply coefficient of correlation), denoted by r , is a measure
of the strength of linear relationship between two variables. The coefficient
                                                                     cov ( x, y )
of correlation between two variables x and y is given by r =                        , where
                                                                       σ xσ y
cov ( x, y ) is covariance of x and y , σ x is the standard deviation for x , σ y is
the standard deviation for y . Correlation describes how the two variables are
related. The correlation coefficient ranges from -1 to +1.
There are three types of correlation:
 • Negative correlation
 • Zero correlation
 • Positive correlation
If the linear coefficient of correlation takes values closer to -1, the correlation is
strong and negative, and will become stronger the closer r approaches -1.
If the linear coefficient of correlation takes values close to 1, the correlation is
strong and positive, and will become stronger the closer r approaches 1. If the
linear coefficient of correlation takes values close to 0, the correlation is weak.
If r = 1 or r=-1, there is perfect correlation and the line on the scatter plot is
increasing or decreasing respectively. If r = 0, there is no linear correlation.
                ∑  yi − ( axi + b )
                                          2
                                                        ∑ ( y − ax − b )
                                                                                2
          D=                                       D=          i        i
That is         i =1                          or        i =1                        (1) is minimum.
  1.   Differentiate relation (1) with respect to b . In this case, x, y and a will
       be considered as constants.
  2.   Equate relation obtained in 1 to zero, divide each side by n and give the
       value of b .
 • Take the value of b obtained in 2 and put it in relation obtained in 1.
   Differentiate the obtained relation with respect to Variance for variable x
                                 2
                1 n 
                  ∑  i 
                             −
   is σ x2
   =                     x − x
                n i =1 
                                                                        2
                                                    1 n          −
                                                                    
                         y is σ y2
 • Variance for variable =                            ∑     yi − y 
                                                    n i =1         
                                                      1 n          −
                                                                             −
                                                                                 
 • Covariance of these two variables is cov ( x, y )=   ∑     xi − x   yi − y  , give
                                                      n i =1                  
   the simplified expression equal to a.
  3.      a , equate it to zero and divide both sides by n to find the value of a .
  4.   Using the relations in 2:
                                                                        cov ( x, y )        − cov ( x, y ) − 
       regression line y on x is written
                                      =  as y                                          x +  y−             x .
                                                                            σ   2
                                                                                x                σ  2
                                                                                                     x        
                  −   cov ( x, y )      −
                                           
  x−x
  =                                 y − y .
                         σ y2             
                                                        −    cov ( x, y )      −
                                                                                  
              This line is written as Lx=
                                        /y ≡ x − x                         y − y .
                                                                σ y2             
    Example
    Find the regression line of y on x for the following data and estimate the value
    of y for x = 4 , x = 7 , x = 16 and the value of x for y = 7 , y = 9 , y = 16 .
     x                  3              5                6               8                          9                  11
     y                  2              3                4               6                          5                  8
     Answer:
         x             y                   −    −                 2                            2
                                     x − x y − y  x − x                                                               
                                                        −                          −                              −          −
                                                     i                         y
                                                                              i − y                       xi − x   yi − y 
                                                                                                                        
     3                 2            -4     -2.6 16                           6.76                          10.4
     5                 3            -2     -1.6 4                            2.56                          3.2
     6                 4            -1     -0.6 1                            0.36                          0.6
     8                 6            1      1.4 1                             1.96                          1.4
     9                 5            2      0.4 4                             0.16                          0.8
     11                8            4      3.4 16                            11.56                         13.6
          6             6                               6         2                            2
                                                                     
         ∑ xi = 42
                                                                    −         6
                       ∑ yi = 28                        ∑                    ∑  y − y 
                                                                                        −                   6
     −          42   − 28
    x
    =           = 7 ,=
                     y = 4.7 ,
                6      6
              1 6      −
                                 −
                                      30
cov ( x, y ) = ∑  xi − x   yi − y  = , = 5
              6 i =1               6
                                2
                1 6         −
                                 42
    σ x2 =        ∑      x
                        i
                6 i =1 
                           − x  = =7 ,
                                 6
     132                   Mathematics | Student Book | Senior Five | Experimental Version
                     2
    1 6            23.36
σ = ∑  yi − y  =
             −
 2
 y                       = 3.89.
    6 i =1          6
                                                      5
 The regression line of y on x is Ly / x ≡ y − 4.7=     ( x − 7) .
                                                      7
x 60 61 62 63 65
       i) Write the death rate for developed countries and the death rate for
          underdeveloped countries in ascending order.
       ii) Rank the death rate for developed countries and the death rate for
           underdeveloped countries such that the lowest death rate is ranked 1
           and the highest death rate is ranked 6.
                                                                       ∑d
                                                                       i =1
                                                                              i
                                                                               2
                                                                                   =?
                                                                                     n
                                                                                   6∑ di2
      iv) Using the information completed in the table above, find 1 −              i =1
                                                                               n ( n − 1)
                                                                                       2
CONTENT SUMMARY
A Spearman coefficient of rank correlation or Spearman’s rho is a measure
of statistical dependence between two variables. It assesses how well the
relationship between two variables can be described using a monotonic
             n
function.6Thed Spearman’s    coefficient of rank correlation is denoted and defined
by
           ∑   i
                2
 ρ = 1 − i =12
        n ( n − 1)
                   , where, d refers to the difference of ranks between paired
items in two series and n is the number of observations.
It is much easier to calculate the Spearman’s coefficient of rank correlation than
to calculate the Pearson’s coefficient of correlation as there is far less working
involved. However, in general, the Pearson’s coefficient of correlation is a more
accurate measure of correlation.
    Income (X in        12      8    16    12    7    10    12     16    12   9
    thousands
    (FRW))
    Expenditure (Y      6       5    7     7     4    6     8      13    10   10
    in thousands
    (FRW))
CONTENT SUMMARY
Bivariate statistics can help in prediction of the value for one variable if we know
the value of the other. Bivariate data occur all the time in real-world situations
and we typically use the following methods to analyze the bivariate data:
  • Scatter plots
  • Correlation Coefficients
  • Simple Linear Regression
The following examples show different scenarios where bivariate data appears
in real life.
Businesses often collect bivariate data about total money spent on advertising
and total revenue. For example, a business may collect the following data for 12
consecutive sales quarters:
She may then create a simple scatter plot to visualize the relationship between
these two variables:
   Price (£10s)             611    811    591     792    520   573    683    716
         a) Calculate the Spearman’s coefficient of rank correlation between:
     i) Price and transport manager’s rankings,
     ii) Price and saleswoman’s grades.
          b) Based on the result of a, state, giving a reason, whether it would
             be necessary to use all the three different methods of assessing
             the cars.
          c) A new employee is asked to collect further data and to do some
             calculations. He produces the following results: The coefficient of
             correlation between
     i) Price and boot capacity is 1.2,
     ii) Maximum speed and fuel consumption in miles per gallons is -0.7,
     iii) Price and engine capacity is -0.9. For each of his results, say giving
         a reason, whether you think it is reasonable.
          d) Suggest two sets of circumstances where Spearman’s coefficient
             of rank correlation would be preferred to the Pearson’s coefficient
             of correlation as a measure of association.
  Student       A         B             C            D             E            F
  Judge1        6.8       7.3           8.1          9.8           7.1          9.2
  Judge2        7.8       9.4           7.9          9.6           8.9          6.9
  Club             A              B     C      D      E       F          G    H
  Position         1              2     3      4      5       6          7    8
  Average position 27             29    9      16     24      15         12   22
 Find both equations of regression lines. Also estimate the value of y for
 x=30.
                       (x)                (y)
 Mean                  53                 142
 Variance              130                165
         n
                  −
                            −
                                
       ∑
       i =1                  
                                  1220
             xi − x   yi − y  =
Find both equations of the regression lines. Also estimate the blood
pressure of a man whose age is 45.